Problem: $y=6x^3\sqrt{x}\csc(x)$ Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-9\csc^{3/2}(x)\cos(x)$ (Choice B) B $-21x^{5/2}\cos(6x^{7/2})\csc(6x^{7/2})$ (Choice C) C $3x^{5/2}(7-2x\cot(x))\csc(x)$ (Choice D) D $-9x^{3/2}\csc(x)\cot(x)$
Answer: $6x^3\sqrt{x}\csc(x)$ is a product of three functions. Let... $u(x)=6x^3$ $v(x)=\sqrt{x}$ $w(x)=\csc(x)$... then $y=u(x)\cdot v(x) \cdot w(x)$. To find $\dfrac{dy}{dx}$, we will need to use the product rule twice! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left[u(x)\cdot v(x) \cdot w(x)\right] \\\\ &=u'(x)\cdot \Bigl(v(x)\cdot w(x) \Bigr)+u(x)\cdot\dfrac{d}{dx}\left[v(x)\cdot w(x)\right]&\gray{\text{Product rule}} \\\\ &=u'(x)\cdot \Bigl(v(x)\cdot w(x) \Bigr)+u(x)\cdot\Bigl(v'(x)\cdot w(x)+v(x)w'(x)\Bigr)&\gray{\text{Product rule}} \\\\ &= u'(x)\cdot v(x) \cdot w(x) + u(x) \cdot v'(x) \cdot w(x) + u(x) \cdot v(x) \cdot w'(x) \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=18x^2$ $v'(x)=\dfrac{1}{2\sqrt{x}}$ $w'(x)=-\csc(x)\cot(x)$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}u'(x)\cdot v(x) \cdot w(x) + u(x) \cdot v'(x) \cdot w(x) + u(x) \cdot v(x) \cdot w'(x) \\\\ &=18x^2 \cdot \sqrt{x} \cdot \csc(x) + 6x^3 \cdot \dfrac{1}{2\sqrt{x}} \cdot \csc(x) + 6x^3 \cdot \sqrt{x} \cdot -\csc(x)\cot(x) \\\\ &=21x^{5/2}\csc(x)-6x^{7/2}\csc(x)\cot(x) \\\\ &=3x^{5/2}(7-2x\cot(x))\csc(x) \end{aligned}$ In conclusion: $\dfrac{dy}{dx}=3x^{5/2}(7-2x\cot(x))\csc(x)$